Let $\alpha: [a,b]\to M$ be an embedded curve in a Riemannian manifold $(M,g)$ and let $p$ be a point in $M$, not on the curve $\alpha$. If $p$ is close enough to $\alpha$, there exists a unique geodesic $\gamma$ of $M$, parametrised proportional to arc length and satisfying $\gamma(0)= \alpha(s)$ for some $s \in [a,b]$, $g(\gamma'(0), \alpha'(s))= 0$ and $p = \gamma(1)$. The image of $p$ under the reflection with respect to $\alpha$ is the point $q = \gamma(-1)$. The $q$ said to be local reflection of $p$.

Now, if we take a non-zero vector field $X$ and use the above algorithm for integral curve of $X$ and suppose local reflection is a isometry, then what we can say about:

Q1: Riemann (or sectional) curvature of the manifolds? Is it completely determined?

Q2: the vector field $X$?

In some Papers

Take $p \in M$ and denote by $f$ the local reflection with respect to the integral curve of $X$ through $p$. Then $f(p) = p$,$f_∗X_p = X_p$ and $f_∗Y_p = -Y_p$ for $Y\in X_p^\perp$ . Since $f$ is a local isometry, it preserves the Riemann curvature tensor. In particular, for $Y_1 ,Y_2 ,Y_3 ∈ T_p M$, it holds

$$f_∗R_p (Y_1 ,Y_2 )Y_3 = R_p (f_∗ Y_ 1 ,f_∗Y_2 )f_∗ Y_3,$$ So, if $Y_1 ,Y_2 ,Y_3$ are orthogonal to $X$, we have $$f_∗R_p (Y_1 ,Y_2 )Y_3 = -R_p (Y_ 1 ,Y_2 )Y_3.$$

In this way, we see that, for $Y,Z,W$ orthogonal to $X$ it holds $$R(Y,Z )W \perp X,\ R(Y,Z )X = 0,\ R(Y,X)Z\sim X,\ R(Y,X)X ⊥ X. $$

Update: I found in some paper the following criterion for a reflection with respect to a curve to be locally isometric:

Theorem: Let $(M,g)$ be a Riemannianm manifold and $\alpha : [a,b]\to M$ an embedded curve in $M$. If the reflection with respect to the curve $\alpha$ is an isometry, then the following hold:

  • $\alpha$ is a geodesic;
  • $g((\nabla^{2k}_{u\cdots u}R)(u, v)u, \alpha^\prime) = 0;$
  • $g((\nabla^{2k+1}_{u\cdots u}R)(u, v)u, w) = 0;$
  • $g((\nabla^{2k+1}_{u\cdots u}R)(u, \alpha^\prime)u, \alpha^\prime) = 0;$

for all vectors $u , v$ and $w$ orthogonal to $\alpha^\prime$ and $k = 0, 1, 2, \cdots$. Moreover, if $M$ is analytic, these conditions are also sufficient for the reflection to be an isometry.


  • 1
    $\begingroup$ A remark and a question. Q: do you want this property for one integral curve of $X$ or for all of them? and R: It seems to me that every curve that produces a local isometry has to be a geodesic. Otherwise, take $\gamma$ pointing in the direction of the geodesic curvature. $\endgroup$ – Sebastian Goette Mar 12 '17 at 19:02
  • 4
    $\begingroup$ First note that all your integral curves have to be geodesics. Moreover, if the sectional curvature of $M$ is constant then then the latter condition is also sufficient. In particular, one can not determine the sign of sectional curvature. More examples come from warped products of symmetric spaces with $\mathbb{R}$ or $\mathbb{S}^1$. $\endgroup$ – Anton Petrunin Mar 12 '17 at 22:57
  • $\begingroup$ for all of integral curve of $X$. $\endgroup$ – C.F.G Mar 13 '17 at 4:45

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